If similarity ratio means the ratio between the lengths of the cubes:
![\frac{ L_1^{3} }{ L_2^{3} } = \frac{V_1}{V_2} \\ \\ ratio = \frac{L_1}{L_2} = \frac{\sqrt[3]{V_1} }{\sqrt[3]{V_2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%20L_1%5E%7B3%7D%20%7D%7B%20L_2%5E%7B3%7D%20%7D%20%3D%20%20%5Cfrac%7BV_1%7D%7BV_2%7D%20%20%5C%5C%20%20%5C%5C%20ratio%20%3D%20%20%5Cfrac%7BL_1%7D%7BL_2%7D%20%3D%20%20%20%5Cfrac%7B%5Csqrt%5B3%5D%7BV_1%7D%20%7D%7B%5Csqrt%5B3%5D%7BV_2%7D%20%7D%20)
A rectangular pyramid is sliced so the cross section is perpendicular to its base but does not pass through its vertex. What is
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I already answered that question, here's the link: brainly.com/question/10534199.
But if you still want the explanation, here it is:
"So if we want to measure probability, you have to add, then divide, like this:
First
, which gives us 18.
Then we see, that 6 is 1/3 of 18, so we conclude. The answer is 33.33%.
Hope this helped!"
But if you still want the explanation, here it is:
"So if we want to measure probability, you have to add, then divide, like this:
First
Then we see, that 6 is 1/3 of 18, so we conclude. The answer is 33.33%.
Hope this helped!"
Answer:
We conclude that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care.
Step-by-step explanation:W
We are given with the cost-to-charge ratios for both inpatient and outpatient care in 2002 for a sample of six hospitals in Oregon below;
Hospital 2002 Inpatient Ratio 2002 Outpatient Ratio
1 68 54
2 100 75
3 71 53
4 74 56
5 100 74
6 83 71
Let = <u><em>mean cost-to-charge ratio for outpatient care</em></u>
= <u><em>mean cost-to-charge ratio for impatient care</em></u>.
SO, Null Hypothesis, :
{means that the mean cost-to-charge ratio for Oregon hospitals is higher or equal for outpatient care than for inpatient care}
Alternate Hypothesis, :
{means that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care}
The test statistics that would be used here is <u>Two-sample t-test statistics</u> because we don't know about population standard deviations;
T.S. = ~
where, = sample mean cost-to-charge Outpatient Ratio =
= 63.83
= sample mean cost-to-charge Impatient Ratio =
= 82.67
= sample standard deviation for Outpatient Ratio =
= 10.53
= sample standard deviation for Impatient Ratio =
= 14.33
= sample of hospital for outpatient care = 6
= sample of hospital for outpatient care = 6
Also, =
= 12.574
So, <u><em>the test statistics</em></u> = ~
= -2.595
The value of t test statistics is -2.595.
<u>Now, at 5% significance level, the t table gives critical value of -1.812 at 10 degree of freedom for left-tailed test.</u>
Since, our test statistics is less than the critical value of t as -2.595 < -1.812, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.
Therefore, we conclude that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care.